What are the solutions to the equation 3y^2-10y=-5 ?

The solutions to the equation 3y^2-10y=-5 are y=(5+sqrt(10))/3 ,y=(5-sqrt(10))/3

Find the zeros of 3y^2-10y=-5

The zeros of 3y^2-10y=-5 are y=(5+sqrt(10))/3 ,y=(5-sqrt(10))/3

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3 y^{2} - 10 y = - 5

y = \frac{5 + 10}{3}, y = \frac{5 - 10}{3}

Decimal Notation

y = 2.72075 \dots, y = 0.61257 \dots

Solution steps

3 y^{2} - 10 y = - 5

Move

5

to the left side

3 y^{2} - 10 y + 5 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 10 ) \pm ( - 10 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

(- 10)^{2} - 4 \cdot 3 \cdot 5 = 210

y_{1, 2} = \frac{- ( - 10 ) \pm 2 10}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- ( - 10 ) + 2 10}{2 \cdot 3}, y_{2} = \frac{- ( - 10 ) - 2 10}{2 \cdot 3}

y = \frac{- ( - 10 ) + 2 10}{2 \cdot 3} : \frac{5 + 10}{3}

y = \frac{- ( - 10 ) - 2 10}{2 \cdot 3} : \frac{5 - 10}{3}

The solutions to the quadratic equation are:

y = \frac{5 + 10}{3}, y = \frac{5 - 10}{3}

3 x^{2} - 4 a x - 2 a^{2} = 0

1 x^{2} - 2 x + 2 = 0

- x^{2} + 144 = 0

2 x^{2} - 3 x + 1

x^{2} + 8 x - 192 = 0

What are the solutions to the equation 3y^2-10y=-5 ?
The solutions to the equation 3y^2-10y=-5 are y=(5+sqrt(10))/3 ,y=(5-sqrt(10))/3
Find the zeros of 3y^2-10y=-5
The zeros of 3y^2-10y=-5 are y=(5+sqrt(10))/3 ,y=(5-sqrt(10))/3